Given the plane curve, find tangent vector

[mdawg467]

Homework Statement

Consider the plane curve \overrightarrow{r(t)}=e^tcost(t)\hat{i}+e^tsin(t) \hat{j}
Find the following when t= ∏/2
Part A: \hat{T}(t)
Part B: \hat{B}(t)
Part C: \hat{N}(t)

Homework Equations

\hat{N}(t)=\frac{\hat{T}(t)}{||\hat{T}(t)||}
\hat{T}(t)=\frac{\overrightarrow{r'(t)}}{|| \overrightarrow{r'(t)}||}

\hat{B(t)}=\frac{\overrightarrow{r'(t)\times r''(t) }}{||\overrightarrow{r'(t)\times r''(t)}||}

The Attempt at a Solution

Part A
\overrightarrow{r(t)}=e^tcost(t)\hat{i}+e^tsin(t) \hat{j}

\overrightarrow{r'(t)}=e^t[(cos(t)-sin(t))\hat{i} \:+\:(sin(t)+cos(t))\hat{j}]

\overrightarrow{r'(\frac{\pi }{2})}=e^\frac{\pi }{2}[(cos(\frac{\pi }{2})-sin(\frac{\pi }{2}))\hat{i} \:+\:(sin(\frac{\pi }{2})+cos(\frac{\pi }{2}))\hat{j}]

\overrightarrow{r'(\frac{\pi }{2})}=-e^\frac{\pi }{2}\hat{i}\;+\;e^\frac{\pi }{2}\hat{j}


\hat{T}(t) =\frac{-e^\frac{\pi }{2}\hat{i}\;+\;e^\frac{\pi }{2}\hat{j}}{ \sqrt{(-e^\frac{\pi }{2})^2\;+\;(e^\frac{\pi }{2})^2} }

Based off of Part A, plugging the numbers into Part B and C generate:
\hat{B(t)}=\frac{\overrightarrow{r'(t)\times r''(t) }}{||\overrightarrow{r'(t)\times r''(t)}||}=0
\hat{N}(t)=\frac{\hat{T}(t)}{||\hat{T}(t)||}=0

Not sure if I solved this correctly.

Any help would be great. Thank you.

 

[Dick]

You got T(t) right. That's about all you've really shown. Your formula for N(t) isn't even right. It's supposed to have a derivative in it.